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EXPAND sqrt{/frac{1-x}{1+x}}
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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\[\Huge \sqrt{\frac{1-x}{1+x}}\]
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up to power x^2
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i get \[1-x+\frac{x^2}{2} - \frac{x^3}{8} - \frac{3x^4}{64}\]

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@saifoo.khan
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wolfram got \[-\frac{3 x^5}{8}+\frac{3 x^4}{8}-\frac{x^3}{2}+\frac{x^2}{2}-x+1\]
saifoo.khan
  • saifoo.khan
Unfortunately i forgot the formula! D:
anonymous
  • anonymous
Is it a maclaurin series?
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you can try that
anonymous
  • anonymous
I want to know what is meant by Expand, since that's the only thing that comes to mind.
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you can use this \[(x+a)^n=\sum ^{n}_{ k=0} \left(\begin{matrix}n \\ k\end{matrix}\right) x^k a^{(n-k)}\]
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or \[(1+x)^n = 1+nx+\frac{(n)(n-1)}{2!}(x^2)+...\]
anonymous
  • anonymous
That works for \(n\in \mathbb{N}\).
anonymous
  • anonymous
In this case you have a square root...
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still can, watch
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\[\sqrt{\frac{1-x}{1+x}} \\ \\ \frac{\sqrt{1-x}}{\sqrt{1+x}} \\ \\ (1-x)^\frac{1}{2}(1+x)^{-\frac{1}{2}}\]
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expand separately
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mix them by multiplying together
anonymous
  • anonymous
Yeah, my problem is with \(1/2 \notin \mathbb{N}\)
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I don't think it should be natural number
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i mean real number will do
anonymous
  • anonymous
How can you have summations or binomial coefficients with rational numbers?
amoodarya
  • amoodarya
interval of x is ?
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the formula list says can use
1 Attachment
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we use the second one (1+x)^n
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http://en.wikipedia.org/wiki/Binomial_series look at the History section
amoodarya
  • amoodarya
|dw:1359798673882:dw|
amoodarya
  • amoodarya
|dw:1359798819059:dw|
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or maybe its \[-\frac{3 x^5}{8}+\frac{3 x^4}{8}-\frac{x^3}{2}+\frac{x^2}{2}-x+1\]
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hmm
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based from the formula as long as its in the from ("1"+x)^n should be fine to expand, i tried it
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from \[(1-x)^{1/2} \\ \\ n=-1/2, x=-x(for formula) \\ \\ [1-\frac{x}{2}+\frac{(\frac{1}{2})(\frac{1}{2}-1)}{2}(x^2)] \\ \\ 1-\frac{x}{2}-\frac{x^2}{8}\]

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